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KonstantinChe [14]
3 years ago
4

The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t

. [5 marks] ii.Does the object ever stop changing? [5 marks] iii.When is the object moving to the right and when is the object moving to the left?
Physics
1 answer:
kozerog [31]3 years ago
0 0

Answer:

1.) V = 12t^3 - 120t^2 + 252t

2.) No

3.) At positive V and negative V

Explanation: Given that the position of an object at any time t is s(t)=3t4−40t3+126t2−9.

1.) To determine the velocity V, we will differenciate the equation above with respect to t.

V = ds/dt

ds/dt = 12t^3 - 120t^2 + 252t

Therefore,

V = 12t^3 - 120t^2 + 252t

2.) The equation above depict an exponential equation which proves that the object never stops changing.

3.) The object moves to the right at positive velocity V and moves to the left at negative velocity V

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What elements are solid liquid and gas at room temperature?
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3 0
2 years ago
Current passes through a solution of sodium chloride. In 1.00 second, 2.68×1016 Na+ ions arrive at the negative electrode and 3.
sladkih [1.3K]

Answer:

10.6 mA

Explanation:

t = time interval = 1.00 s

q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C

n₁ = number of Na⁺ ions = 2.68 x 10¹⁶

q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C

n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶

q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C

i₁ = Current due to Na⁺ ions = \frac{q_{1}}{t} = \frac{0.004288}{1} = 0.004288 A

i₂ = Current due to Cl⁻ ions = \frac{q_{2}}{t} = \frac{0.006272}{1} = 0.006272 A

Current passing between the electrodes is given as

i = i₁ + i₂

i = 0.004288 + 0.006272

i = 0.01056 A

i = 10.6 x 10⁻³ A

i = 10.6 mA

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